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March 17th, 2018, 05:00 AM   #1
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Projectile motion


A man is standing on a railway car which moves at a constant velocity of 5m/s He releases an object in his hand from a height of 1.8m as shown in the figure; at the same instant the man releases the object, the railway car start slowing down at a deceleration of 1m/s^2. At what horizontal distance from point L on the car does the object strike the car?
Please explain step by step.

Last edited by skipjack; March 17th, 2018 at 11:52 PM.
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March 17th, 2018, 06:29 AM   #2
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time it takes for the object to hit the surface of the cart ...

$-1.8 = -4.9t^2 \implies t = \dfrac{3\sqrt{2}}{7}$


horizontal displacement of the object ...

$x_1 = 5t$

horizontal displacement of point L ...

$x_2 = 5t - \dfrac{t^2}{2}$

can you finish?
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March 17th, 2018, 09:45 PM   #3
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Quote:
Originally Posted by skeeter View Post
time it takes for the object to hit the surface of the cart ...

$-1.8 = -4.9t^2 \implies t = \dfrac{3\sqrt{2}}{7}$


horizontal displacement of the object ...

$x_1 = 5t$

horizontal displacement of point L ...

$x_2 = 5t - \dfrac{t^2}{2}$

can you finish?
The object is freely falling where did the horizontal displacement of object come from?
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March 18th, 2018, 03:45 AM   #4
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Free fall does not mean an object moves strictly in a vertical direction. It will do so only if it has a zero velocity component in the horizontal direction.

$x = x_0 + v_{0x} t - \dfrac{1}{2}a_x t^2$

Both object and position L on the cart’s surface start at $x_0=0$ and both move with an initial horizontal velocity $v_{0x}= 5 \, m/s$

The object’s horizontal velocity remains a constant 5 m/s. It undergoes no acceleration in the horizontal direction after release.

$x_1 = 0 + 5t - 0$

The cart’s initial horizontal velocity is 5 m/s, but it slows due to a constant acceleration opposite its initial direction of motion.

$x_2 = 0 + 5t + \dfrac{1}{2}(-1) t^2$

projectile motion on a moving cart
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Last edited by skeeter; March 18th, 2018 at 04:16 AM.
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